Answer
$135^{o},\quad 225^{o}$
Work Step by Step
Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$
$\left[\begin{array}{lllll}
Quad.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
Cosine is negative in quadrants II and III.
Browsing through Function Values of Special Angles,
we find that $\displaystyle \cos 45^{o}=\frac{\sqrt{2}}{2}$, so the reference angle is $45^{o}$
In quadrant II, $\theta^{\prime}=180^{o}-\theta$ so
$\theta=180^{o}-\theta^{\prime}=180^{o}-45^{o}=135^{o}$
In quadrant III, $\theta^{\prime}=\theta-180^{o}$ so
$\theta=180^{o}+\theta^{\prime}=180^{o}+45^{o}=225^{o}$
